Problem: Simplify and expand the following expression: $ \dfrac{p}{p + 1}-\dfrac{p + 6}{2p - 1} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(p + 1)(2p - 1)$ Multiply the first term by $\dfrac{2p - 1}{2p - 1}$ $ \begin{align*} \dfrac{p}{p + 1} \times \dfrac{2p - 1}{2p - 1} & = \dfrac{(p)(2p - 1)}{(p + 1)(2p - 1)} \\ & = \dfrac{2p^2 - p}{(p + 1)(2p - 1)}\end{align*} $ Multiply the second term by $\dfrac{p + 1}{p + 1}$ $ \begin{align*} \dfrac{p + 6}{2p - 1} \times \dfrac{p + 1}{p + 1} & = \dfrac{(p + 6)(p + 1)}{(2p - 1)(p + 1)} \\ & = \dfrac{p^2 + 7p + 6}{(2p - 1)(p + 1)}\end{align*} $ Now we have: $ = \dfrac{2p^2 - p}{(p + 1)(2p - 1)} - \dfrac{p^2 + 7p + 6}{(2p - 1)(p + 1)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{2p^2 - p - (p^2 + 7p + 6)}{(p + 1)(2p - 1)} $ $ = \dfrac{2p^2 - p - p^2 - 7p - 6}{(p + 1)(2p - 1)} $ $ = \dfrac{p^2 - 8p - 6}{(p + 1)(2p - 1)}$ Expand the denominator: $ = \dfrac{p^2 - 8p - 6}{2p^2 + p - 1}$